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  If the cosmos is expanding, what is the effect on $1/r^2$?

+ 1 like - 0 dislike

Space in the universe is expanding, and this effect can be only detected at large scales, on the size of clusters of galaxies and larger.

If space is expanding, the expression $1/r^2$ for the law of gravity will be modified. Is this correct? At what scales and how exactly would this occur? 

Put differently, if two masses are really distant, say 1% or 10% of the Hubble Radius, or even more, does $1/r^2$ still hold? Is there a simple way to answer? Is gravity increased or decreased? By how much?

Does general relativity exclude such an effect?

asked May 15, 2020 in Astronomy by Gina [ no revision ]
recategorized May 16, 2020 by Dilaton

Why are you asking about $1/r^2$ and not just about distances $\Delta r$ and periods $\Delta t$?

2 Answers

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Yes the law of gravity is modified. Before a more detailed answer can be given, the problem has to be clarified.

You refer to the gravitational effects between two masses. So let us consider two masses in an otherwise empty space. Let us further assume one mass to be "large" and the other mass to be a "test particle". Then, according to general relativity, you have to solve the Einstein equations for the large mass. Assuming that it is spherically symmetric and non-rotating (and not charged) you arrive at the Schwarzschild solution for the spacetime metric. From this you can find an effective potential for the motion of the test particle. You have a deviation from the Newtonian effective potential for small values of \(r\).

You also refer to the expansion of the universe. The simplest modification to take this into account in the above setup is to consider the Einstein equation with cosmological constant. Solving for the spacetime metric and finding the effective potential will now yield an additional deviation for large \(r\).

Qualitatively it could be said that for not too large \(r\) the expansion of spacetime is mitigated by the large mass, but for larger \(r\) the expansion wins. The relevant length scale here is \(1/\sqrt\Lambda\), with \(\Lambda\) the cosmological constant.

answered May 15, 2020 by Flamma (90 points) [ no revision ]

So, if there is a cosmological constant, is Newton's law $1/r^2$ weakened for the test mass?

Even without cosmological constant Newton's law is not the full story, as I indicated above. See for example Chandrasekhar "The Mathematical Theory of Black Holes", Chapter 3. With cosmological constant, you still have an attractive term \(\propto 1/r\) in the effective potential (i.e. "Newton"), but you also have a repulsive term \(\propto r^2/\lambda^2\), where \(\lambda\) is the length scale in the answer above, and is in the Gpc range.

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It would be modified to ȑ=-GM/r²+Λc²r/3, so at a distance of r=³√(GM/H²)=³√(3GM/Λ/c²) the test particle would keep a constant distance to the dominant mass M if it is initially at rest. Here we assume a universe with a hubble constant H=c√(Λ/3) where Λ is the cosmological constant and a dominant mass M, so the metric is described by the Schwarzschild De Sitter metric.

answered Jul 21, 2020 by Yukterez (10 points) [ revision history ]
edited Jul 23, 2020 by Yukterez

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